### SOLUTIONS TO CONTINUITY OF FUNCTIONS OF ONE VARIABLE

SOLUTION 1 : Function f is defined at x=1 since

i.) f(1) = 2 .

The limit

= 3 (1) - 5

= -2 ,

i.e.,

ii.) .

But

iii.) ,

so condition iii.) is not satisfied and function f is NOT continuous at x=1 .

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SOLUTION 2 : Function f is defined at x=-2 since

i.) f(-2) = (-2)2 + 2(-2) = 4-4 = 0 .

The left-hand limit

= (-2)2 + 2(-2)

= 4 - 4

= 0 .

The right-hand limit

= (-2)3 - 6(-2)

= -8 + 12

= 4 .

Since the left- and right-hand limits are not equal, ,

ii.) does not exist,

and condition ii.) is not satisfied. Thus, function f is NOT continuous at x=-2 .

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SOLUTION 3 : Function f is defined at x=0 since

i.) f(0) = 2 .

The left-hand limit

= 2 .

The right-hand limit

= 2 .

Thus, exists with

ii.) .

Since

iii.) ,

all three conditions are satisfied, and f is continuous at x=0 .

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SOLUTION 4 : Function h is not defined at x=-1 since it leads to division by zero. Thus,

i.) h(-1)

does not exist, condition i.) is violated, and function h is NOT continuous at x = -1 .

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SOLUTION 5 : First, check for continuity at x=3 . Function f is defined at x=3 since

i.) .

The limit

(Circumvent this indeterminate form by factoring the numerator and the denominator.)

(Recall that A2 - B2 = (A-B)(A+B) and A3 - B3 = (A-B)(A2+AB+B2 ) . )

(Divide out a factor of (x-3) . )

=

,

i.e.,

ii.) .

Since,

iii.) ,

all three conditions are satisfied, and f is continuous at x=3 . Now, check for continuity at x=-3 . Function f is not defined at x = -3 because of division by zero. Thus,

i.) f(-3)

does not exist, condition i.) is violated, and f is NOT continuous at x=-3 .

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SOLUTION 6 : Functions y = x2 + 3x + 5 and y = x2 + 3x - 4 are continuous for all values of x since both are polynomials. Thus, the quotient of these two functions, , is continuous for all values of x where the denominator, y = x2 + 3x - 4 = (x-1)(x+4) , does NOT equal zero. Since (x-1)(x+4) = 0 for x=1 and x=-4 , function f is continuous for all values of x EXCEPT x=1 and x=-4 .

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SOLUTION 7 : First describe function g using functional composition. Let f(x) = x1/3 , , and k(x) = x20 + 5 . Function k is continuous for all values of x since it is a polynomial, and functions f and h are well-known to be continuous for all values of x . Thus, the functional compositions

and

are continuous for all values of x . Since

,

function g is continuous for all values of x .

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SOLUTION 8 : First describe function f using functional composition. Let g(x) = x2 - 2x and . Function g is continuous for all values of x since it is a polynomial, and function h is well-known to be continuous for . Since g(x) = x2 - 2x = x(x-2) , it follows easily that for and . Thus, the functional composition

is continuous for and . Since

,

function f is continuous for and .

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SOLUTION 9 : First describe function f using functional composition. Let and . Since g is the quotient of polynomials y = x-1 and y = x+2 , function g is continuous for all values of x EXCEPT where x+2 = 0 , i.e., EXCEPT for x = -2 . Function h is well-known to be continuous for x > 0 . Since , it follows easily that g(x) > 0 for x < -2 and x > 1 . Thus, the functional composition

is continuous for x < -2 and x > 1 . Since

,

function f is continuous for x < -2 and x > 1 .

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SOLUTION 10 : First describe function f using functional composition. Let and h(x) = e x , both of which are well-known to be continuous for all values of x . Thus, the numerator is continuous (the functional composition of continuous functions) for all values of x . Now consider the denominator . Let g(x) = 4 , h(x) = x2 - 9 , and . Functions g and h are continuous for all values of x since both are polynomials, and it is well-known that function k is continuous for . Since h(x) = x2 - 9 = (x-3)(x+3) = 0 when x=3 or x=-3 , it follows easily that for and , so that is continuous (the functional composition of continuous functions) for and . Thus, the denominator is continuous (the difference of continuous functions) for and . There is one other important consideration. We must insure that the DENOMINATOR IS NEVER ZERO. If

then

.

Squaring both sides, we get

16 = x2 - 9

so that

x2 = 25

when

x = 5 or x = -5 .

Thus, the denominator is zero if x = 5 or x = -5 . Summarizing, the quotient of these continuous functions, , is continuous for and , but NOT for x = 5 and x = -5 .

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SOLUTION 11 : Consider separately the three component functions which determine f . Function is continuous for x > 1 since it is the quotient of continuous functions and the denominator is never zero. Function y = 5 -3x is continuous for since it is a polynomial. Function is continuous for x < -2 since it is the quotient of continuous functions and the denominator is never zero. Now check for continuity of f where the three components are joined together, i.e., check for continuity at x=1 and x=-2 . For x = 1 function f is defined since

i.) f(1) = 5 - 3(1) = 2 .

The right-hand limit

=

(Circumvent this indeterminate form one of two ways. Either factor the numerator as the difference of squares, or multiply by the conjugate of the denominator over itself.)

= 2 .

The left-hand limit

=

= 5 - 3(1)

= 2 .

Thus,

ii.) .

Since

iii.) ,

all three conditions are satisfied, and function f is continuous at x=1 . Now check for continuity at x=-2 . Function f is defined at x=-2 since

i.) f(-2) = 5 - 3(-2) = 11 .

The right-hand limit

=

= 5 - 3( -2)

= 11 .

The left-hand limit

=

= -1 .

Since the left- and right-hand limits are different,

ii.) does NOT exist,

condition ii.) is violated, and function f is NOT continuous at x=-2 . Summarizing, function f is continuous for all values of x EXCEPT x=-2 .

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SOLUTION 12 : First, consider separately the two components which determine function f . Function y = A2 x - A is continuous for for any value of A since it is a polynomial. Function y = 4 is continuous for x < 3 since it is a polynomial. Now determine A so that function f is continuous at x=3 . Function f must be defined at x=3 , so

i.) f(3)= A2 (3) - A = 3 A2 - A .

The right-hand limit

=

= A2 (3) - A

= 3 A2 - A .

The left-hand limit

=

= 4 .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

so that

3 A2 - A - 4 = 0 .

Factoring, we get

(3A - 4)(A + 1) = 0

for

or A = -1 .

For either choice of A ,

iii.) ,

all three conditions are satisfied, and f is continuous at x=3 . Therefore, function f is continuous for all values of x if or A = -1 .

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SOLUTION 13 : First, consider separately the three components which determine function f . Function y = Ax - B is continuous for for any values of A and B since it is a polynomial. Function y = 2x2 + 3Ax + B is continuous for for any values of A and B since it is a polynomial. Function y = 4 is continuous for x > 1 since it is a polynomial. Now determine A and B so that function f is continuous at x=-1 and x=1 . First, consider continuity at x=-1 . Function f must be defined at x=-1 , so

i.) f(-1)= A(-1) - B = - A - B .

The left-hand limit

=

= A (-1) - B

= - A - B .

The right-hand limit

=

= 2(-1)2 + 3A(-1) + B

= 2 - 3A + B .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

so that

2A - 2B = 2 ,

or

(Equation 1)

A - B = 1 .

Now consider continuity at x=1 . Function f must be defined at x=1 , so

i.) f(1)= 2(1)2 + 3A(1) + B = 2 + 3A + B .

The left-hand limit

=

= 2(1)2 + 3A(1) + B

= 2 + 3A + B .

The right-hand limit

=

= 4 .

For the limit to exist, the right- and left-hand limits must exist and be equal. Thus,

ii.) ,

or

(Equation 2)

3A + B = 2 .

Now solve Equations 1 and 2 simultaneously. Thus,

A - B = 1 and 3A + B = 2

are equivalent to

A = B + 1 and 3A + B = 2 .

Use the first equation to substitute into the second, getting

3 (B + 1 ) + B = 2 ,

3 B + 3 + B = 2 ,

and

4 B = -1 .

Thus,

and

.

For this choice of A and B it can easily be shown that

iii.)

and

iii.) ,

so that all three conditions are satisfied at both x=1 and x=-1 , and function f is continuous at both x=1 and x=-1 . Therefore, function f is continuous for all values of x if and .

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SOLUTION 14 : First describe f using functional composition. Let g(x) = -1/x2 and h(x) = ex . Function h is well-known to be continuous for all values of x . Function g is the quotient of functions continuous for all values of x , and is therefore continuous for all values of x except x=0 , that x which makes the denominator zero. Thus, for all values of x except x=0 ,

f(x) = h ( g(x) ) = e g(x) = e -1/x2

is a continuous function (the functional composition of continuous functions). Now check for continuity of f at x=0 . Function f is defined at x=0 since

i.) f(0) = 0 .

The limit

(The numerator approaches -1 and the denominator is a positive number approaching zero.)

,

so that

= 0 ,

i.e.,

ii.) .

Since

iii.) ,

all three conditions are satisfied, and f is continuous at x=0 . Thus, f is continuous for all values of x .

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SOLUTION 15 : First show that f is continuous for all values of x . Describe f using functional composition. Let , , and k(x) = x2 . Function h is well-known to be continuous for all values of x . Function k is a polynomial and is therefore continuous for all values of x . Function g is the quotient of functions continuous for all values of x , and is therefore continuous for all values of x except x=0 , that x which makes the denominator zero. Thus, for all values of x except x=0 ,

is a continuous function (the product and functional composition of continuous functions). Now check for continuity of f at x=0 . Function f is defined at x=0 since

i.) f(0) = 0 .

The limit does not exist since the values of oscillate between -1 and +1 as x approaches zero. However, for

so that

.

Since

,

it follows from the Squeeze Principle that

ii.) .

Since

iii.) ,

all three conditions are satisfied, and f is continuous at x=0 . Thus, f is continuous for all values of x . Now show that f is differentiable for all values of x . For we can differentiate f using the product rule and the chain rule. That is, for the derivative of f is

.

Use the limit definition of the derivative to differentiate f at x=0 . Then

.

Use the Squeeze Principle to evaluate this limit. For

.

If , then

.

If , then

.

In either case,

,

and it follows from the Squeeze Principle that

.

Thus, f is differentiable for all values of x . Check to see if f' is continuous at x=0 . The function f' is defined at x=0 since

i.) f'(0) = 0 .

However,

ii.)

does not exist since the values of oscillate between -1 and +1 as x approaches zero. Thus, condition ii.) is violated, and the derivative , f' , is not continuous at x=0 .

NOTE : The continuity of function f for all values of x also follows from the fact that f is differentiable for all values of x .

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Duane Kouba
1998-06-01